Question

A metallic sphere floats in an immiscible mixture of water (${\mathrm{\rho }}_{\mathrm{W}}={10}^3\mathrm{kg}/{\mathrm{m}}^3$) and a liquid (${\mathrm{\rho }}_{\mathrm{L}}=13.5\times {10}^3$) such that its ${\left(\frac{4}{5}\right)}^{\mathrm{th}}$volume is in water and ${\left(\frac{1}{5}\right)}^{\mathrm{th}}$volume is in the liquid. Find the density of the metal.

• A. $4.5\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$
• B. ${\text{4.0×10}}^3\mathrm{kg}/{\mathrm{m}}^3$
• C. $3.5\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$
• D. $1.9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$

Answer 1

The correct option is C

$3.5\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$

Step 1: Given parameters

The density of water, ${\mathrm{\rho }}_{\mathrm{W}}={10}^3\mathrm{kg}/{\mathrm{m}}^3$

The density of the liquid, ${\mathrm{\rho }}_{\mathrm{L}}=13.5\times {10}^3$

Volume in water ${\left(\frac{4}{5}\right)}^{\mathrm{th}}$

Volume in liquid ${\left(\frac{1}{5}\right)}^{\mathrm{th}}$

Step 2: To find

We have to find the density of the metal.

Step 3: Calculation

We know that,

$\mathrm{Total}\mathrm{upthrust}=\mathrm{Weight}\mathrm{of}\mathrm{the}\mathrm{metal}\mathrm{sphere}$

Hence,

$\left(\frac{4}{5}\mathrm{V}\right){\mathrm{\rho }}_{\mathrm{W}}\mathrm{g}+\left(\frac{1}{5}\mathrm{V}\right){\mathrm{\rho }}_{\mathrm{L}}\mathrm{g}=\left({\mathrm{V\rho }}_{\mathrm{m}}\right)\mathrm{g}$

Here,

${\mathrm{\rho }}_{\mathrm{W}}$ is the density of water.

${\mathrm{\rho }}_{\mathrm{L}}$ is the density of the liquid.

$\mathrm{V}$ is the volume.

${\mathrm{\rho }}_{\mathrm{m}}$ is the density of the metal.

By substituting the value,

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{m}}=\left(\frac{4}{5}\right)\times {10}^3+\left(\frac{1}{5}\right)\times 13.5\times {10}^3 \\ =3.5\times {10}^3\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Therefore, option B is the correct choice.